What is the extraneous solution to these equations? $\dfrac{x^2 + 2}{x + 3} = \dfrac{-x + 8}{x + 3}$
Answer: Multiply both sides by $x + 3$ $ \dfrac{x^2 + 2}{x + 3} (x + 3) = \dfrac{-x + 8}{x + 3} (x + 3)$ $ x^2 + 2 = -x + 8$ Subtract $-x + 8$ from both sides: $ x^2 + 2 - (-x + 8) = -x + 8 - (-x + 8)$ $ x^2 + 2 + x - 8 = 0$ $ x^2 - 6 + x = 0$ Factor the expression: $ (x + 3)(x - 2) = 0$ Therefore $x = -3$ or $x = 2$ At $x = -3$ , the denominator of the original expression is 0. Since the expression is undefined at $x = -3$, it is an extraneous solution.